Difference Equation Explained by Oppenheim

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SUMMARY

The discussion centers on the homogeneous difference equation presented in "Discrete-Time Signal Processing" by Oppenheim, specifically the equation \(\sum_{k=0}^N a_k y_h[n-k] = 0\). The solutions to this equation are expressed as \(y_h[n] = \sum_{m=1}^N A_m z^n_m\), indicating that the sequence \(y_h[n]\) belongs to a family of solutions characterized by complex numbers \(z_m\). These complex numbers are identified as the roots of the polynomial \(\sum_{k=0}^N a_k z^{-k} = 0\), which is directly related to the z-transform.

PREREQUISITES
  • Understanding of homogeneous difference equations
  • Familiarity with the z-transform
  • Knowledge of polynomial roots
  • Basic concepts from "Discrete-Time Signal Processing" by Oppenheim
NEXT STEPS
  • Study the properties of homogeneous difference equations
  • Learn about the z-transform and its applications in signal processing
  • Explore polynomial root-finding techniques
  • Review the solutions to differential equations for comparative analysis
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Students and professionals in signal processing, mathematicians, and engineers seeking to deepen their understanding of difference equations and their solutions as presented by Oppenheim.

wildman
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This is some math from "Discrete-Time Signal Processing" by Oppenheim:

We have the homogeneous difference equation:

[tex]\sum_{k=0}^N a_k y_h [n-k] = 0[/tex]

"The sequence [tex]y_h[n][/tex] is in fact a member of a family of solutions of the form:

[tex]y_h[n] = \sum_{m=1}^ N A_m z^n_m[/tex]"

So what is Oppenheim saying here? I suppose it is something like the solutions to differential equations. Right? But what is the z? I suppose it is related to the z transform, right?

He then says: Substituting the second equation for the first shows that the complex numbers [tex]z_m[/tex] must be roots of the polynomial:

[tex]\sum_{k=0}^N a_k z^{-k} = 0[/tex]

Could some one explain this to me?

Thanks!
 
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i think this might be what you're looking for
http://tutorial.math.lamar.edu/AllBrowsers/3401/SeriesSolutions.asp
 
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